a, \(Fe+H_2SO_{4\text{loãng}}\rightarrow FeSO_4+H_2\)

\(n_{Fe}=n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(Fe+H_2SO_{4\text{đặc}}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+H_2O\)

\(Cu+H_2SO_{4\text{đặc}}\rightarrow CuSO_4+SO_2+H_2O\)

Bảo toàn e:

\(2n_{Cu}+3n_{Fe}=2n_{SO_2}\)

\(\Leftrightarrow n_{Cu}=\dfrac{2n_{SO_2}-3n_{Fe}}{2}=0,25\left(mol\right)\)

\(\Rightarrow x=m_{Cu}+m_{Fe}=0,25.64+0,5.56=44\left(g\right)\)

a) Đặt \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=b=n_{Fe}\\n_{SO_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\end{matrix}\right.\)

Bảo toàn electron: \(2a+3b=2\) \(\Rightarrow2a+3\cdot0,5=2\) \(\Rightarrow a=n_{Cu}=0,25\left(mol\right)\)

\(\Rightarrow x=m_{Cu}+m_{Fe}=0,25\cdot64+0,5\cdot56=44\left(g\right)\)

b) Ta có: \(n_{H_2SO_4\left(p/ư\right)}=\dfrac{1}{2}n_{e\left(traođổi\right)}+n_{SO_2}=\dfrac{1}{2}\cdot2+1=2\left(mol\right)\)

\(\Rightarrow\Sigma n_{H_2SO_4\left(đặc\right)}=2\cdot110\%=2,2\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{2,2\cdot98}{98\%}=220\left(g\right)\) \(\Rightarrow V_{H_2SO_4}=\dfrac{220}{1,84}\approx119,57\left(ml\right)\)

c) Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=1\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,4\cdot1,5=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối

PTHH: \(2SO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HSO_3\right)_2\)

              2x                 x                    x          (mol)

            \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3\downarrow+H_2O\)

              y             y                                      (mol)

Ta lập được hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,6\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=n_{Ba\left(HSO_3\right)_2}=0,4\left(mol\right)\\y=0,2\end{matrix}\right.\)

\(\Rightarrow C_{M_{Ba\left(HSO_3\right)_2}}=\dfrac{0,4}{0,4}=1\left(M\right)\)

1)

- Xét phần 1:

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,2<-------------------0,2

=> n_Fe = 0,2 (mol)

- Xét phần 2:

\(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

             0,2-->0,6-------->0,1--------->0,3

            Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

            0,3<----0,6<------0,3<-----0,3

=> n_Cu = 0,3 (mol)

m = 2.(0,2.56 + 0,3.64) = 60,8 (g)

2)

\(m_{H_2SO_4\left(bđ\right)}=\dfrac{200.98}{100}=196\left(g\right)\)

=> \(m_{H_2SO_4\left(sau.pư\right)}=196-98\left(0,6+0,6\right)=78,4\left(g\right)\)

m_{dd sau pư} = \(\dfrac{60,8}{2}+200-0,6.64=192\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1.400}{192}.100\%=20,83\%\\C\%_{\left(CuSO_4\right)}=\dfrac{0,3.160}{192}.100\%=25\%\\C\%_{\left(H_2SO_4.dư\right)}=\dfrac{78,4}{192}.100\%=40,83\%\end{matrix}\right.\)

Đáp án : B

Nếu NaOH dư => n F e O H 3  = 0,2 mol

=> n_Fe(bd) = 0,2 mol

=> n_O(bd) = 0,5 mol. Vô lý

(Vì số mol O trong oxit cao nhất là gấp 1,5 lần số mol Fe).

Vậy NaOH thiếu : 19,2  F e : a O : b

56a + 16b = 19,2

n H 2 S O 4 = 0 , 9

→ B T D T n S O 4 2 - = 0 , 3 + 3 a 2

→ B T N T . S n S O 2 = 0 , 9 - 0 , 3 + 3 a 2 a

6a – 2b = 1,5 ⇒ a = 0 , 3 b = 0 , 15 → V = 6 , 72

Đáp án A

Đáp án C